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nikitadnepr [17]
3 years ago
13

An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fl

uid flow through the 3.0-inch section will be what factor times that through the 1.0-inch section
Engineering
1 answer:
geniusboy [140]3 years ago
5 0

Answer:

(\frac{r_1}{r_2})^2=\frac{1}{9}

Explanation:

From the question we are told that:

Diameter 1 d_1=1.0

Diameter 2 d_2=3.0

Generally the equation for Radius is mathematically given by

At Diameter 1

r_{1}=\frac{1}{2} inch

At Diameter 2

r_{2}=\frac{3}{2} inch

Generally the equation for continuity is mathematically given by

 A_1V_1=A_2V_2

Therefore

(\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2

(\frac{r_1}{r_2})^2=\frac{1}{9}

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Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Nutka1998 [239]

Answer:

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3 years ago
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
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Answer:

#include <iostream>

using namespace std;

void PrintPopcornTime(int bagOunces) {

if(bagOunces < 3){

 cout << "Too small";

 cout << endl;

}

else if(bagOunces > 10){

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}

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int main() {

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Using C++ to write the program. In line 1 we define the header "#include <iostream>"  that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.

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